Practicing Success
Practicing Success
CUET Preparation Today
If $(x-\frac{1}{x})^2= 12$, what is the value of $(x^2-\frac{1}{x^2})$, given that x > 0? |
$6\sqrt{2}$ $8\sqrt{3}$ $6\sqrt{3}$ $8\sqrt{2}$ |
$8\sqrt{3}$ |
If $(x-\frac{1}{x})^2= 12$ then, $(x-\frac{1}{x})$ = \(\sqrt {12}\) = 2\(\sqrt {3}\) and $(x+\frac{1}{x})$ = \(\sqrt {12 + 4}\) = 4 The value of $(x^2-\frac{1}{x^2})$ = 4 × 2\(\sqrt {3}\) = $8\sqrt{3}$ |
